Question

If x=1+a+a^2+a^3....to infinity;

and y=1+b+b^2+b^3+...to infinity

Then prove that 1+ab+a^2b^2+...to infinity=xy/(x+y-1)

Answer 1

x=1+a+a2+.....∞

Clearly , the above expression is a G.P (geometric progression)

And sum of infinite term of a G.P. (a+ar+ar2+....∞ )is given by

S(∞)= a/1−r,

where a is the first term and r is the common ratio of the G.P.

Now ,

x=1+a+a2.......∞

=> x = 1/1−a

=> 1-a = 1/x

=> a= 1−1/x……………(1)

y= 1+b+b2+.....∞

=> y = 1/1−b

=> 1-b=1/y

=> b= 1−1/y……………(2)

Now,

1+ab+a2b2+........∞

= 1/1−ab

= 1/1−(1−1/x)(1−1/y)

= 1/1−(1+1/xy−1/x-1/y)

= 1/1/x+1/y−1/xy

=xy/x+y−1